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Chapter 4: Problem 12

Find (a) the partial derivatives \(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\) and \(\frac{\partial f}{\partial z}\) and (b) the gradient \(\nabla f(x, y, z)\) $$ f(x, y, z)=\frac{x+y}{y+z} $$

Short Answer

Expert verified
The partial derivatives are \( \frac{1}{y+z}, \frac{z-x}{(y+z)^2}, \text{and} \ \frac{-(x+y)}{(y+z)^2} \), and the gradient is \( \left( \frac{1}{y+z}, \frac{z-x}{(y+z)^2}, \frac{-(x+y)}{(y+z)^2} \right) \).

Step by step solution

01

Find Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) and \( z \) as constants. Differentiating the function \( f(x, y, z) = \frac{x+y}{y+z} \) with respect to \( x \), we use the fact that a constant with respect to \( x \) vanishes. Thus, \( \frac{\partial}{\partial x}(x+y) = 1 \) and the denominator \( y+z \) remains unchanged, giving us \( \frac{\partial f}{\partial x} = \frac{1}{y+z} \).
02

Find Partial Derivative with Respect to y

For \( \frac{\partial f}{\partial y} \), apply the quotient rule \( \frac{d}{dy}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dy} - u \frac{dv}{dy}}{v^2} \). Here, \( u = x + y \) and \( v = y + z \). Hence \( \frac{du}{dy} = 1 \) and \( \frac{dv}{dy} = 1 \). Compute:- \( v \frac{du}{dy} = (y+z) \cdot 1 = y + z \).- \( u \frac{dv}{dy} = (x+y) \cdot 1 = x + y \).The partial derivative is: \( \frac{(y+z) - (x+y)}{(y+z)^2} = \frac{z-x}{(y+z)^2} \).
03

Find Partial Derivative with Respect to z

Use the quotient rule again for \( \frac{\partial f}{\partial z} \) where \( \frac{du}{dz} = 0 \) since \( x+y \) is independent of \( z \), and \( \frac{dv}{dz} = 1 \) since \( v = y + z \). Applying the quotient rule gives:- \( (y+z) \cdot 0 = 0 \).- \( (x+y) \cdot 1 = x + y \).Then, \( \frac{\partial f}{\partial z} = \frac{0 - (x+y)}{(y+z)^2} = \frac{-(x+y)}{(y+z)^2} \).
04

Compute the Gradient

The gradient of \( f \), denoted as \( abla f(x, y, z) \), is the vector comprising the partial derivatives:\[ abla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = \left( \frac{1}{y+z}, \frac{z-x}{(y+z)^2}, \frac{-(x+y)}{(y+z)^2} \right) \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Gradient in Multivariable Calculus
The concept of a gradient is pivotal in multivariable calculus. In simple terms, the gradient is a vector that represents both the direction and rate of the steepest ascent of a function.
It's essential for functions of multiple variables, giving us critical insights into their behavior. When you have a function like \(f(x, y, z)\), the gradient \(abla f\) is compiled from its partial derivatives with respect to each variable. Each partial derivative tells us how the function changes as only one of the variables is varied.
For example, with our function \(f(x, y, z) = \frac{x+y}{y+z}\), the gradient would be:
  • \(\frac{\partial f}{\partial x}\)
  • \(\frac{\partial f}{\partial y}\)
  • \(\frac{\partial f}{\partial z}\)

All packed into a vector. This vector tells us the direction in which \(f\) increases most rapidly, and its magnitude tells us how fast \(f\) increases in that direction.
Insights into Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of more than one variable.
This branch is crucial for understanding and analyzing phenomena in higher dimensions. In such cases, instead of dealing with simple curves, we handle surfaces and higher-dimensional constructs. Partial derivatives play a vital role here.
They give us precise aspects of how a function changes as one of its inputs varies while keeping the others constant.
  • They are the building blocks for understanding more complex transformations and calculations.

  • Understanding each partial derivative's significance helps in appreciating the whole landscape of the function's behavior.
A handy tool in this area is the gradient we discussed earlier. It aids in dealing with spatial changes effectively.
Mastering the Quotient Rule
The quotient rule is an essential tool when dealing with derivatives in calculus, especially when tackling functions expressed as the ratio of two other functions.
This rule helps find the derivative of something like \(f(x) = \frac{u}{v}\), where both \(u\) and \(v\) are functions of some variable. The quotient rule states that: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] Applying this, as shown in the solution, one can derive partial derivatives concerning different variables. Understanding and using this rule can simplify many problems in calculus. It is fundamental for solving situations where division of functions is involved, such as rationing rates or slopes in multivariable contexts.

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Most popular questions from this chapter

Let \(S\) be the ellipsoid \(x^{2}+\frac{y^{2}}{2}+\frac{z^{2}}{4}=4 .\) Let \(\mathbf{a}=(1,2,2),\) and let \(\mathbf{n}\) be the unit normal vector to \(S\) at a that points outward from \(S\). If \(f(x, y, z)=x y z,\) find the directional derivative \(\left(D_{\mathbf{n}} f\right)(\mathbf{a})\)

Consider the problem of finding the maximum and minimum values of the function \(f(x, y)=\) \(x y\) subject to the constraint \(x+2 y=1\) (a) Explain why a minimum value does not exist. (b) On the other hand, you may assume that a maximum does exist. Use Lagrange multipliers to find the maximum value and the point at which it is attained.

Let \(f(x, y, z)=x^{2} y+y^{2} z+z^{2} x,\) and let \(\mathbf{a}=(1,-1,1)\). (a) Find \(\frac{\partial f}{\partial x}(\mathbf{a}), \frac{\partial f}{\partial y}(\mathbf{a}),\) and \(\frac{\partial f}{\partial z}(\mathbf{a})\) (b) Find \(D f(\mathbf{a})\) and \(\nabla f(\mathbf{a})\). (c) Find the first-order approximation \(\ell(x, y, z)\) of \(f(x, y, z)\) at a. (You may assume that \(f\) is differentiable at a.) (d) Compare the values of \(f(1.05,-1.1,0.95)\) and \(\ell(1.05,-1.1,0.95)\).

Let \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) be the function: $$ f(x, y)=\sqrt{x^{4}+y^{4}} $$ (a) Find formulas for the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) at all points \((x, y)\) other than the origin. (b) Find the values of the partial derivatives \(\frac{\partial f}{\partial x}(0,0)\) and \(\frac{\partial f}{\partial y}(0,0)\). (Note that the formulas you found in part (a) probably do not apply at \((0,0) .)\) (c) Use the definition of differentiability to determine whether \(f\) is differentiable at (0,0) .

Find (a) the partial derivatives \(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\) and \(\frac{\partial f}{\partial z}\) and (b) the gradient \(\nabla f(x, y, z)\) $$ f(x, y, z)=x y+x z+y z-x y z $$
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